mjhjmtu

I came across a perplexing thing while testing some assumptions in Wolfram|Alpha, and as I don't have a math education beyond college algebra, I thought this would be a good place to ask. I would just like to emphasize that this is not a homework question, and is purely curiosity.

Common faulty math proofs tend to use logic like this:

1^0 = 1
2^0 = 1
therefore 1 = 2

This is obviously false because x^0 is defined as 1 for all real values of x.

But in the case of e^(i2π) = e^0, the base is the same: e. So if a^x = a^y is true, shouldn't it follow, then, that x = y and therefore i2π = 0? If not, why?

That's because the exponential isn't an injective function on $mathbb C$. If is an injective function on $mathbb R$, and so the single-valued logarithm is well-defined on $mathbb R^+$ but not on $mathbb C$.

This isn't true for the same reason that $(-1)^2=1^2$ doesn't imply that $-1=1$.

Recall that

$$f(x)=f(y) implies x=y$$

require that $f$ is an injective function.

Simply because the exponential function is periodic with period $2pi i$.

That means that for any $x$, the values of $e^x$ and $e^x+2pi i k$ are equal for all integers $k$.

In particular, you are looking at $x=0$ and $k=0,1$; so $e^0=e^2pi i$, but of course $0$ and $2pi i$ are different numbers.